Question

If $5$ distinct balls are placed at random into $5$ cells, then the probability that exactly one cell remains empty is

• A. $\frac{48}{125}$
• B. $\frac{12}{125}$
• C. $\frac{8}{125}$
• D. $\frac{1}{125}$

Answer 1

The correct option is A $\frac{48}{125}$

Since, we have to keep one cell empty,
therefore one of the cells must have two balls.
Two balls can be chosen in ${}^5{C}_2$ ways
The cell containing two balls can be choosen in ${}^5{C}_1$ ways
Remaining $3$ balls can be arranged in $4$ cells in number of ways $={}^4{P}_3=4!$

Total number of ways$={}^5{C}_2\times {}^5{C}_1\times 4!=1200$
Required Probability $=\frac{1200}{5^5}=\frac{48}{125}$